Screw Gauge And Micrometer
Measuring a wire, a screw gauge of least count 0.01 mm shows a main scale reading of 2 mm and a circular scale reading of 45 divisions, with a zero error of plus 3 divisions.
Select the correct option:
Solution
2.42 mm
As explained in NCERT Class 11, Chapter 2, the corrected reading of a screw gauge is the observed reading minus the zero error, where the observed reading is the main scale reading plus the circular scale division count times the least count. First compute the observed reading: 2 mm+(45×0.01 mm)=2+0.45=2.45 mm. A positive zero error means the instrument reads too high, so it must be subtracted: zero error =+3×0.01=+0.03 mm. The corrected diameter is 2.45−0.03=2.42 mm. The value 2.45 mm is wrong because it ignores the zero error entirely, reporting only the observed reading. The value 2.48 mm is wrong because it adds the zero error instead of subtracting a positive one, moving in the wrong direction. The value 2.03 mm is wrong because it forgets to convert the 45 circular divisions with the least count, adding only the zero correction. The sign rule is worth stressing: a positive zero error means the circular scale zero lies above the reference line when the jaws are closed, so the instrument reads high and the error is subtracted, whereas a negative zero error would be added back. To avoid crushing the wire, the ratchet at the end of the thimble should be used so that the same gentle pressure is applied every time, keeping readings reproducible. A sanity check confirms that a positive zero error correctly reduces the measured diameter, and the final 2.42 mm is a physically reasonable thin-wire thickness.
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About This Question
- Subject
- physics
- Chapter
- experimental skills
- Topic
- screw gauge and micrometer
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.42 mm
As explained in NCERT Class 11, Chapter 2, the corrected reading of a screw gauge is the observed reading minus the zero error, where the observed reading is the main scale reading plus the circular scale division count times the least count. First compute the observed reading: 2 mm+(45×0.01 mm)=2+0.45=2.45 mm. A positive zero error means the instrument reads too high, so it must be subtracted: zero error =+3×0.01=+0.03 mm. The corrected diameter is 2.45−0.03=2.42 mm. The value 2.45 mm is wrong because it ignores the zero error entirely, reporting only the observed reading. The value 2.48 mm is wrong because it adds the zero error instead of subtracting a positive one, moving in the wrong direction. The value 2.03 mm is wrong because it forgets to convert the 45 circular divisions with the least count, adding only the zero correction. The sign rule is worth stressing: a positive zero error means the circular scale zero lies above the reference line when the jaws are closed, so the instrument reads high and the error is subtracted, whereas a negative zero error would be added back. To avoid crushing the wire, the ratchet at the end of the thimble should be used so that the same gentle pressure is applied every time, keeping readings reproducible. A sanity check confirms that a positive zero error correctly reduces the measured diameter, and the final 2.42 mm is a physically reasonable thin-wire thickness.
This medium difficulty physics question is from the chapter experimental skills, covering the topic of screw gauge and micrometer. It appeared in the 2025 exam.
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